Simplify; express your answer in exponential form. Assume $k\neq 0, p\neq 0$. $\dfrac{{(k^{-3}p)^{4}}}{{(k^{-4}p^{-2})^{-1}}}$
Solution: To start, try simplifying the numerator and the denominator independently. In the numerator, we can use the distributive property of exponents. ${(k^{-3}p)^{4} = (k^{-3})^{4}(p)^{4}}$ On the left, we have ${k^{-3}}$ to the exponent ${4}$ . Now ${-3 \times 4 = -12}$ , so ${(k^{-3})^{4} = k^{-12}}$ Apply the ideas above to simplify the equation. $\dfrac{{(k^{-3}p)^{4}}}{{(k^{-4}p^{-2})^{-1}}} = \dfrac{{k^{-12}p^{4}}}{{k^{4}p^{2}}}$ Break up the equation by variable and simplify. $\dfrac{{k^{-12}p^{4}}}{{k^{4}p^{2}}} = \dfrac{{k^{-12}}}{{k^{4}}} \cdot \dfrac{{p^{4}}}{{p^{2}}} = k^{{-12} - {4}} \cdot p^{{4} - {2}} = k^{-16}p^{2}$